Calculating Normal & Friction Forces for 1350kg Car on Plane Surface

  • Thread starter Thread starter fibersnet
  • Start date Start date
  • Tags Tags
    Forces
Click For Summary

Homework Help Overview

The problem involves calculating the normal and friction forces exerted by a 1350kg car resting on a plane surface, with additional factors such as a cable tension and direction cosines influencing the forces at play.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up force equations based on the car's weight, cable tension, and the normal force, but expresses confusion over the resulting normal force calculation.
  • Some participants question the interpretation of the frictional vector components and their relation to the overall force equations.

Discussion Status

The discussion has seen some productive exchanges, with the original poster indicating they resolved their confusion regarding the frictional vector components. However, there remains a lack of explicit consensus on the correct values for the forces involved.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific requirements on the approach and solutions. The original poster's calculations suggest a discrepancy in expected results, indicating potential misunderstandings in the setup or assumptions made.

fibersnet
Messages
3
Reaction score
0
1350kg car resting on a plane surface with its brakes locked.
Unit vector [tex]e_n = <.231,.923,.308>[/tex] is perpendicular to the surface. The y-axis points upwards. The direction cosines of a cable supporting the car are [tex]<-.816,.408,-.408>[/tex] and the tension in the cable is 1.2KN. Determine the magnitude of the normal and friction forces the car's wheels exert on the surface.

This is what I have so far:

[tex]F_f = <.231F_x, .923F_y, .308F_z>[/tex]
[tex]T_{AB} = 1200<-.816, .408, -.408>[/tex]
[tex]W = -(1350)*(9.8)\hat{j}[/tex]
[tex]N = |1200|<.231,.923,.308>[/tex]

[tex]0 = F_f + T_{AB} + N - W[/tex]


But when I solve the equations, I come up with N being 9719N, it is supposed to be about 2500 more.

Thanks!
 
Physics news on Phys.org
Picture

Here is a picture of the diagram and the math.
 

Attachments

  • car.jpg
    car.jpg
    6.8 KB · Views: 490
  • math.gif
    math.gif
    5.5 KB · Views: 523
Last edited:
.231N + X -.816*1200=0
.923N+Y-1350*9.8+1200*.408=0
.308N + Z -.408*1200=0
The fourth is correct.
 
Got it

Thanks, I figured this out late last night right before i went to bed.

I forgot that <fx,fy,fz> were already in the form of the frictional vector and that the equation .231Fx + .923Fy + .308Fz was just meant to relate the components, they are not the actual components.

Thanks!
 

Similar threads

Replies
11
Views
3K
Replies
18
Views
3K
Replies
3
Views
2K
Replies
8
Views
5K
  • · Replies 20 ·
Replies
20
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
1
Views
10K
Replies
7
Views
2K